Cantor Confusion
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From: mueckenh on 13 Dec 2006 16:06 William Hughes schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > William Hughes schrieb: > > > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > > William Hughes schrieb: > > > > > > > > > > > A_1 = {1} > > > > > > > A_2 = {1,2} > > > > > > > A_3 = {1,2,3} > > > > > > > > > > > > > > B = {1,2,3} > > > > > > > > > > > > > > then B is contained in the last A_i. If there is no last A_I, then > > > > > > > there is > > > > > > > no A_i that contains B > > > > > > > > > > > > That has nothing to do with "last". > > > > > > > > > > If A_i contains B, then A_i contains any A_j. > > > > > Therefore A_i is "last". > > > > > > > > > > > No comment? > > > > If B contains any A_i then B is the last. > > B cannot be the last A_i, as B is not > one of the A's (B corresponds to the diagnonal, > the A_i correspond to the lines) In which element does it differ from every A_i? Don't forget, the A_i form a linear set. It is not possible for B to differ from A_n in element a but not in element b and from A_m in element b but not in element a. This approach has been entertaind without a thought being given to it. But it is wrong. Regards, WM
From: mueckenh on 13 Dec 2006 16:18 cbrown(a)cbrownsystems.com schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Dik T. Winter schrieb: > > > > > In article <1165923504.410525.226600(a)79g2000cws.googlegroups.com> mueckenh(a)rz.fh-augsburg.de writes: > > > > Dik T. Winter schrieb: > > > ... > > > > > So you can not show a bijection (in your opinion), but nevertheless you > > > > > state that you have given a surjection. Do you not think you are > > > > > contradicting yourself a bit? > > > > > > > > I give a surjection on all existing paths. > > > > > > But not from the edges. But from parts of edges. > > > > > > > > > No? Even if there are enough parts to gather more than a whole edge to > > > > > > be mapped on every path? > > > > > > > > > > No. Each part of an edge may map to a single path, that does *not* give > > > > > a map from each edge to a single path. > > > > > > > > Since infinity = infinity became an allowed equation, 1/2 + 1/2 = 1 can > > > > no longer be true? > > > > > > So if the first half of ab edge maps to, say, 1/2 and the other half to, > > > say, 3/4. To what single path maps the edge itself? > > > > Why isn't it sufficient to collect the shares of two edegs for every > > path? That shows that there are not more paths than edges. Can you > > explain your objection to factions? > > > > > I previously asked you to "show" how you perform this mapping, given > that we agree to the following: > > (1) Edges are countable. > (2) Paths can be bijected with the reals. > (3) B(e) is the (uncountable) set of all paths to which edge e belongs. > (4) Shares (of edges in the original diagram) are denoted (p,e) for > some path p and some edge e where e belongs to p. > (5) S(e) is the (uncountable) set of all shares of edge e. > (6) C(p) is the (countable) set of all shares which belong to path p. > (7) 1 + 1/2 + 1/4 + ... +1/2^n + ... = 2 > > Your claim (which you seem unable to completely articulate) Don't mix up lacking understanding of he matter with lacking precision in stating it. > seems to be > that there are two "full edges" whose "shares" are exactly the shares > of edges composing a particular path p. But a path p is composed of a > countable number of shares; so a "full edge" can have at most a > countable number of shares. You are aware of the number 2^omega being a countable one? If so, why then do you speak of an uncountable set of shares? Do you think that lim {n-->oo}1/2^n * 2^n consists of uncountably many parts? > Meanwhile, an edge in the original diagram > consists of an uncountable number of shares. So full edges can only be > said to be "the same as" edges in the original diagram if you /assume/ > that countable is the same as uncountable. > > But that is what you are trying to prove - you can't simply assume it > in a proof of it. How can 1 + 1/2 + 1/4 + ... +1/2^n + ... yield the value 2 if we add only countably many shares which are uncountably small? Would you mind to think about that topic again? Now let's argue with aleph0 (in modern set theory it is the same as omega, although there is a big difference). The problem with Cantor's theorem aleph0 < 2^aleph0 is that the function f(x) = 2^x must be discontinuous: For natural x = n we have 2^n < aleph0, for the least infinite x = aleph0 we have 2^aleph > alep0. So there is a gap. The values between aleph0 and 2^aleph0 cannot be taken by the function 2^x. This is highly suspicious, if aleph0 is to be a number which is in trichotomy with the natural numbers, but I can't prove it wrong. By means of the binary tree I can exclude any discontinuity. Therefore the number of branches isolated at a certain level n grows continuosly from 1 to its maximum value. If this was 2^aleph0, then the number of branches could not but call on aleph0 or at least at a value between aleph0 and 2^aleph0. To show that this is impossible is achieved by the big advantage of the tree: its undoubted continuity. Regards, WM
From: mueckenh on 13 Dec 2006 16:27 cbrown(a)cbrownsystems.com schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > Dik T. Winter schrieb: > > > > > > > > > > That one which goes left from the root is not engaged, because we need > > > > > > only half of the set of edges. > > > > > > > > > > So your construction is not a surjection. > > > > > > > > Of course it is! It is a surjection from the set of edges onto the set > > > > of paths. > > > > > > Than again. To what number maps the edge that goes left from the root? > > > If it is a surjection, there should be one. > > > > > A surjection onto the paths covers all elements of the *range* = every > > edge. > > What? The *range* of a surjection from edges onto paths is /not/ every > edge; it is every path. The /domain/ of the surjection is every edge. > (You actually teach Analysis, and you don't know this?!?!) Did you never copy and paste a wrong piece? I corrected it already: The range is "every path". > > > It is not necessary that every edge is mapped on a path, to show > > that there are not less edges than paths. It is only necessary that > > there is no path without edge mapped on it. > > > > That is poorly stated. The mapping that maps the first edge to the left > to every path provides us with a mapping such that "there is no path > without edge mapped to it"; but this mapping does not prove that the > cardinality of the set of paths <= the cardinality ({a single edge}) = > 1. You misunderstood. What I explained to Dik is another mapping than that discussed by us. It is a mapping of one single edge on the path representing 1/3, the next edge is mapped on the path representing pi, and so on. I claim that for every real number I can name an edge to be mapped on this real number. And noboy can disprove it by constructing a diagonal number, because the tree contains them all. Regards, WM
From: mueckenh on 13 Dec 2006 16:38 Franziska Neugebauer schrieb: > mueckenh(a)rz.fh-augsburg.de wrote: > > > Franziska Neugebauer schrieb: > >> mueckenh(a)rz.fh-augsburg.de wrote: > >> > Franziska Neugebauer schrieb: > >> >> mueckenh(a)rz.fh-augsburg.de wrote: > >> >> > Franziska Neugebauer schrieb: > >> >> >> mueckenh(a)rz.fh-augsburg.de wrote: > >> >> >> > Dik T. Winter schrieb: > >> >> >> [...] > >> >> >> >> Again you have provided neither a definition of "number", > >> >> >> >> nor of "grow". > >> >> >> >> Are you unable to do so? In common parlance, but that is > >> >> >> >> not mathematics. In mathematics functions can grow in > >> >> >> >> relation to their argument, but not the entities they > >> >> >> >> denote. > >> >> >> > > >> >> >> > Functions cannot grow, according to modern mathematics. > >> >> >> > >> >> >> Wrong. I have provided a definition: > >> >> >> > >> >> >> ,----[ > >> >> >> <45742128$0$97220$892e7fe2(a)authen.yellow.readfreenews.net> ] > >> >> >> | Definition: A function f: A |-> B grows iff there exist a1 < > >> >> >> | a2 of dom(f) and f(a1) < f(a2). We use the abbreviation "f > >> >> >> | grows" for of "the function f grows". > >> >> >> `---- > >> >> > > >> >> > The natural number n is a particular set of n elements. > >> >> > If the number n can take a value n_1 and can take a value n_2 > >> >> > with n_1 =/= n_2 then the number n can vary. > >> >> > >> >> How is that related to your sentence that "functions cannot grow"? > >> > > >> > It is the usual set theoretic view and as such in direct > >> > opposistion to my view. > >> > >> Can't see any support for your claim, that "functions cannot grow". > > > > In *set theory* a set does not grow. In set theory a function is a > set. In set theory a function cannot grow. > > _IN_ Z set theory there is no _notion_ of /growing/ in the first place. > Without _defining_ what "to grow" shall mean sets and function do > neither grow nor not grow. Without a definition of "to grow" it is just > _meaningless_ to speak of "growing" sets or of sets/functions which > "can" or "cannot grow". > > Hence my definition, which enables us to meaningfully speak of "growing" > functions (but not of sets in general) in order to model in particular > "political" functions like the number of states of the EC at given time. > > > In my view a function can grow. > > Without a proper _definition_ of "to grow" it is meaningless to Z set > theoretically write about function which "can grow". Which _definition_ > do you have in mind? > > > A function can be abbreviated as number variable or as variable > > number. > > Could you explain, what "to abbreviate" means in the present context? > Yes, but I'm not inclined to do it. Good night. WM
From: David R Tribble on 13 Dec 2006 17:11
William Hughes schrieb: >> Do you now claim >> the natural numbers do not exist? > Mueckenh wrote: > More than enoug do exist. (More than we will ever need could be brought > to existence.) How is a number "brought into existence"? I'm thinking of a natural number, call it q. It has the distinct property that it is not equal to any number you (MH) can think of. In fact, it is larger than any natural number that has been "brought into existence yet". So what is q? |